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Correct answer is (b) unheard

The idea that being good can be unpleasant appealed to the children. The word ‘horrible’ in connection with goodness was a novelty. It seemed to introduce a ring of truth that was absent from the aunt’s tales of infant life.

They saw a large windmill on the hill in front of them.

Let one angle be x. It is given that other angle is one-third of first.

So, other angle will be x/3.

Again, given that both the angles are making a linear pair.

So, their sum will be 180°

x + x/3 = 180°

(3x + x)/3 = 180°

4x/3 = 180°

x = (180° x 3)/4

x = 135°

Hence, the angles are 135° and 1/3 x 135°, i.e. 135° and 45°.

1. The absence of sheep in the Prince’s park: There were no sheep in the Prince’s park because his mother had a dream once that her son would be killed by either a sheep or a clock falling on him. 

2. The absence of flowers in the Prince’s park: There were no flowers in the Prince’s park because the pigs had eaten them all. 

3. Bertha feeling sorry for the absence of flowers in the Prince’s park: Bertha had promised her aunts that she would not pick any of the kind Prince’s flowers and she had meant to keep her promise. It made her feel silly to find there were no flowers to pick.

Sancho’s shouts and warnings were not heard by Don Quixote. He flew towards the windmill and collided with it.

When a wolf entered the park and started chasing her, Bertha began to wish that she had never been allowed into the park.

The descriptive element, the unexpected twist, convincing answers to the questions raised in the course of the story made the children conclude that it was the best story that they had ever heard.

The bachelor was annoyed and he stared at the aunt twice and at the communication cord once. It suggested that he was planning to pull it. So, the aunt wanted to pacify the children by telling them a story.

Bertha was happy about being extraordinarily good because she was the only girl who was permitted to get into the Prince’s park and take a walk once a week. But later when she saw a wolf stealing towards her, she was nervous and felt she shouldn’t have entered the Prince’s park. She felt that if she had not been extraordinarily good, she would have been safe in the town.

Bertha won several medals for goodness. She earned a medal for obedience, another medal for punctuality and a third for good behaviour.

Points: 

A, B, C, D, E, G 

Line segments :

\(\overline {AB}, \overline {AD}, \overline {AE}, \overline {BDDE},\overline {BC}, \overline {CD}, \overline {BE}\)

Rays :

\(\overrightarrow {BA}, \overrightarrow {DA}, \overrightarrow {BE},\overrightarrow {AE}, \overrightarrow {EA}, \overrightarrow {DE},\)

Lines :

AE

From the figure 35° + 105° + y = 180° 

∴ y = 180° – 140° = 40° 

∴ x = 40° (∵ x, y are corresponding angles)

He told his squire Sancho that the ogre (windmill) was more seriously wounded than him.

The relationship between Don Duixote and Sancho Panza is an important one. Readers can easily understand that the two characters stand for different things. Don Quixote represents illusion. On the other hand Sancho Panza represents reality. They complement each other in a dualistic way Sancho Panza was a peasant labourer. He was greedy, but kind and faithful. He was also a coward. He is a foil to Don Quixote and virtually to every other character in the story. Don Quixote sees what his mind and imagination create, not which is actually there. He retreats to a world that holds meaning for him. He wants to be a knight in search of his own adventures, winning fame and honour. Unlike Don, Sancho is more practical. He seeks fortune and has a lot of common sense. But he consistently defers with his master and assents to dangerous schemes. However, Don Quixote despite his folly, Superior to the real world he has to deal with

As Don Quixote imagined himself placed in the world of knights, he didn’t believe Sancho.

He was irritated by the noise made by the children and the fact that the aunt was unable to control them.

The wolf would have gone away without spotting Bertha and would have probably satiated its hunger by eating a piglet.

From the given figure we know that

AB and CD are parallel line and BC is a transversal

We know that ∠BCD and ∠ABC are alternate angles

So we can write it as

∠BCD + ∠ABC = x^o

We also know that BC || ED and CD is a transversal

From the figure we know that ∠BCD and ∠EDC form a linear pair of angles

So it can be written as

∠BCD + ∠EDC = 180^o

By substituting the values we get

∠BCD + 75^o = 180^o

On further calculation we get

∠BCD = 180^o – 75^o

By subtraction

∠BCD = 105^o

From the figure we know that ∠BCD and ∠ABC are vertically opposite angles

So we get

∠BCD = ∠ABC = x = 105^o

∠ABC = x = 105^o

Therefore, the value of x is 105^o.

Bertha was so good that she won several medals for her goodness. There were medals for obedience, for good behavior and for punctuality. Ultimately, when news spread and the Prince of the country heard about it, he even allowed her to walk in his beautiful park once a week. This was an opportunity which no other child got.

Don Quixote set spurs to his horse Rozinante and charged into the midst of the sheep and lambs. At this, the frightened animals fled helter-skelter. The shepherds seeing the cause of their disorder, pelted stones at Don Quixote. He soon fell wounded to the ground. All the local peasants thought that Don Quixote was crazy. As he passed, they laughed and insulted him. Don thought that they were praising him. But Sancho told him that they were only mocking at him.

It was a beautiful park. There were no flowers in the park but there were lots of pigs running all over. They were of different colours. There were ponds with colourful fish in them. The fish were gold, blue and green coloured. The trees were full of birds such as the humming bird which hummed the popular tunes of the day and the beautiful parrots that said clever things at a moment’s notice.

Sancho continued his journey with Don Quixote after the windmill incident because he wanted to prevent Don Quixote from attempting anything careless

He thinks that the neighing of horses, the sound of trumpets and the rattling of drums would lead to a battle between the two armies.

The aunt said that the sheep were being driven to a field where there was more grass.

He thought that they were a great army marching towards them. So he started to name the leaders and the principal knights in each army and described different nations.

One consisted of a small girl, a smaller girl, a small boy and their aunt. The other was a bachelor.

As Don Quixote and Sancho pass a farm, they see a thick cloud of dust. Don Quixote thinks it to be a great army marching towards them.

By smacking the cushions of the seat.

1. and
2. but
3. so
4. unless
5. or
6. or
7. nor
8. or
9. but
10. Neither, nor

It is given that

∫ sec⁴ x dx

We can write it as

= ∫sec² x sec² x dx

So we get

= ∫ (1 + tan² x) sec² x dx

Take tan x = t

By differentiation we get

sec² x dx = dt

It can be written as

= ∫(1 + t²) dt

By integrating w.r.t. t

= t + t³/3 + c

By substituting the value of t

= tan x + tan³ x/3 + c

It is given that

∫sin³ x cos x dx

By taking sin x = t

We get

cos x = dt/dx

It can be written as

cos x dx = dt

So we get

= ∫t³ dt

By integrating w.r.t t

= t⁴/4 + c

By substituting the value of t

= sin⁴ x/4 + c

Take x² + 1 = t

So we get

2x dx = dt

By integrating w.r.t. t

∫sin t dt = – cos t + c

By substituting the value of t

= – cos (x² + 1) + c

Take log (sin x) = t

So we get

cos x/sin x dx = dt

By cross multiplication

cot x dx = dt

By integrating w.r.t. t

∫1/t dt = log t + c

By substituting the value of t

= log (log sin x) + c

Take cos² x = t

So we get

– sin 2x dx = dt

By integrating w.r.t. t

∫- e^t dt = – e^t + c

By substituting the value of t

= -e cos² x + c

Take tan x = t

So we get

sec² x dx = dt

By integrating w.r.t. t

∫e^t dt = e^t + c

By substituting the value of t

= e^{tan x} + c

Take √x = t

So we get

1/2√x dx = dt

By integrating w.r.t. t

∫cos t 2 dt = 2 sin t + c

By substituting the value of t

= 2 sin √x + c

No, everybody is not fortunate enough to have a cosy bed to lie in when it rains. Not everybody gets to enjoy the comfort of cosy homes during rain. I have seen animals seeking shelter under trees and under the tin roofs of the small roadside tea stalls.

The people passing by shoo away these animals and try to shrink themselves under the limited space of these shops. The poor animals are left shivering and drenching on the roads. The shopkeepers of such stalls are delighted as the people waiting for the rain to subside often end up buying tea and snacks.

Given : 

Temperature in degree Fahrenheit = 50°F

To find : 

Temperature in degree Celsius

Formula :

°F = \(\frac{9}{5}\)(°C) + 32

Calculation : 

Substituting 50°F in the formula,

°F = \(\frac{9}{5}\)(°C) + 32

50 =  \(\frac{9}{5}\)(°C) + 32

°C = \(\frac{(50-32)\times 5}{9}\) 

= 10°C

ii. Given : 

Temperature in degree Fahrenheit = 10°F

To find : 

Temperature in degree Celsius

Formula : 

°F = \(\frac{9}{5}\)(°C) + 32

Calculation : 

Substituting 10°F in the formula,

°F = \(\frac{9}{5}\)(°C) + 32

10 =  \(\frac{9}{5}\)(°C) + 32

°C = \(\frac{(10-32)\times 5}{9}\) 

= -12.2°C

∴ i. The temperature 50°F corresponds to 10°C.

ii. The temperature 10°F corresponds to -12.2°C.

i. The relationship between degree Fahrenheit and degree Celsius is expressed as,

°F = \(\frac{9}{5}\)(°C) + 32

ii. The relationship between Kelvin and degree Celsius is expressed as,

K = °C + 273.15

P(x) = 9x⁹ – 4x⁸ + 4x⁷ – 3x⁶ + 2x⁵ + x³ + 7x² + 7x + 2

The number of sign changes in P(x) is 4. 

∴ P(x) has atmost 4 positive roots. 

P(-x) = -9x⁹ – 4x⁸ - 4x⁷ – 3x⁶ - 2x⁵ - x³ + 7x² - 7x + 2

The number of sign changes in P(-x) is 3. 

Therefore, P(x) has atmost 3 negative roots.

Let I = ∫x²e^x dx

= x²∫e^x dx - ∫{(dx²/dx) ∫x²e^x} dx

= x²e^x - ∫2x e^x dx

= x²e^x - 2∫xe^x dx

= x²e^x - 2[x∫e^x dx -  ∫(dx/dx) x (e^x) dx

= x²e^x - 2[xe^x - ∫e^x] dx

= x²e^x - 2[xe^x - e^x] +  c

= x²e^x - 2xe^x - e^x + c

Given : 

The circumference of a circle exceeds the diameter by 16.8 cm.

To find : 

The circumference of the circle.

Solution : 

Let diameter of circle = x cm 

So, 

acc. to given condition 

Circumference = x+16.8 cm 

Circumference of circle is 2πr. 

⇒ 2πr = x + 16.8 

Diameter = 2r

⇒ \(\frac{22}7\times{x} = x + 16.8\) 

⇒ \(\frac{22}7{x} -x = 16.8\)

⇒ \(\frac{22x - 7x}7= 16.8\)

⇒ \(\frac{15x}7= 16.8\)

⇒ 15 x = 16.8 × 7

⇒15 x = 117.6

⇒ x = \(\frac{117.6}{15}\)

⇒ x = 7.84

Circumference= x + 16.8 (x = 2r) 

Circumference = 7.84 + 16.8 = 24.64 cm

The perimeter of a circle is called the circumference of the circle. Using a rape or thread we measure the circumference of a wheel or a bangle. We can also measure the diameter. Diameter is the largest chord of a circle. In each case divide the circumference by the diameter. What do you notice? We find that in each case, the ratio of the circumference to the diameter is the same. This ratio is a constant called π (pi) which is a Greek letter.

π= Circumference of circle/Diameter

Form this we get the circumference of a circle is the product of π and its diameter.

C = πd

= 2πr (since d = 2r)

π is a Greek letter. It is a ratio of the circumference to diameter of a circle. Many mathematicians have given many values for this ratio. In the chapter Real numbers, we saw that ' π ' is an irrational number. Aryabhatta gave the value of π as 62832/20000 approximately, whereas Ramanujan found the value of π correct to a million places of decimals. However, for practical purposes, we use π as 22/7. But 22/7 is a rational number. It is only an approximate value.

(a) DE × DC

We know that, area of parallelogram = base × corresponding height

Then, area of parallelogram ABCD = DC × BF

AD × BE = BC × BE … [because AD = BC]

Given AB = 8 cm; BC = 10 cm

Distance from AB and BC is 9 cm

Area of quadrilateral is given as base × height 

= 8 cm × 9 cm 

= 72 cm²

We have, 

f(x) = (x – 1)(x + 2)² 

Differentiate w.r.t x, we get, 

f ‘(x) = (x + 2)² + 2(x – 1)(x + 2) 

= (x + 2)(x + 2 + 2x – 2) 

= (x + 2)(3x) 

For all maxima and minima, 

f ’(x) = 0 

= (x + 2)(3x) = 0 

= x = 0, – 2 

At x = – 2 

f ’(x) changes from –ve to + ve 

Since, 

x = – 2 is a point of Maxima 

At x = 0 

f ‘(x) changes from –ve to + ve 

Since, 

x = 0 is point of Minima. 

Hence, local min value = f(0) = – 4 

local max value = f( – 2) = 0.

Given, as f(x) = x³ – 6x² + 9x + 15

On differentiating with respect to x, we get, f‘(x) = 3x² – 12x + 9 = 3(x² – 4x + 3)

= 3 (x – 3) (x – 1)

For all the maxima and minima,

f’(x) = 0

= 3(x – 3) (x – 1) = 0

= x = 3, 1

At x = 1 f ’(x) changes from negative to positive

Since, x = – 1 is a point of Maxima

At x =3 f‘(x) changes from negative to positive

Since, x =3 is point of Minima.

Thus, local maxima value f (1) = (1)³ – 6(1)² + 9(1) + 15 = 19

Local minima value f (3) = (3)³ – 6(3)² + 9(3) + 15 = 15

Given, as f(x) = (x – 1) (x + 2)²

On differentiating with respect to x, we get,

f‘(x) = (x + 2)² + 2(x – 1)(x + 2)

= (x + 2) (x + 2 + 2x – 2)

= (x + 2) (3x)

For all the maxima and minima,

f’(x) = 0

= (x + 2) (3x) = 0

On solving the above equation we get

= x =0, – 2

At x = – 2 f’(x) changes from negative to positive

Since, x = – 2 is a point of Maxima

At x =0 f‘(x) changes from negative to positive

Since, x = 0 is point of Minima.

Thus, local min value = f (0) = – 4

Local max value = f (– 2) = 0.